If $H$ is a group of order $6$ and $f:S_n\to H$ surjective homomorphism, then what is $n$?

Question

Let $H$ be a group of order $6$, and let $f:S_n\to H$ surjective homomorphism which is not injective.

What is $n$?

My try;

I know that $\ker f$ contains all elements of order not divisible by $2,3$. I wanted to show that for $n\gt 4$, the number of such elements is greater than $|\ker f|=n!/6$ but got stuck trying to prove it by induction (not even sure if it's true). Also, I failed finding homomorphism between $S_4$ to $H$.

For $n\geq 5$, the only normal subgroups of $S_n$ are the identity subgroup, the alternating subgroup $A_n$, and the whole $S_n$ itself. Also $S_4$ has a normal subgroup of order $4$. — Batominovski, Feb 02 '20 at 02:28

score 1 · Accepted Answer · 2020-02-02T07:02:15.473

1

$n$ has to be greater than $3$, since $|S_n|\gt6$. But $n\lt5$ since $S_n$ has only $A_n$ as a normal subgroup for $n\ge 5$. Therefore $n=4$.

edited Feb 02 '20 at 07:02

answered Feb 02 '20 at 06:49

If $H$ is a group of order $6$ and $f:S_n\to H$ surjective homomorphism, then what is $n$?

1 Answers1